AP Chemistry FRQ

Question

PbI2(s)⇄Pb2+(aq)+2I−(aq)     Ksp=7×10−9

The dissolution of PbI2(s) is represented above.

(a) Write a mathematical expression that can be used to determine the value of S , the molar solubility of PbI2(s) .

(b) If PbI2(s) is dissolved in 1.0MNaI(aq), is the maximum possible concentration of Pb2+(aq) in the solution greater than, less than, or equal to the concentration of Pb2+(aq) in the solution in part (a) ? Explain.

Compound	Ksp
PbCl2	2×10−5
PbI2	7×10−9
Pb(IO3)2	3×10−13

c) A table showing Ksp values for several lead compounds is given above. A saturated solution of which of the compounds has the greatest molar concentration of Pb2+(aq) ? Explain.

Answer 1

(a) Ksp = [Pb²⁺][I^-]²

7x10^-9 = (s)(2s)² = 4s³

(b) There is a common ion effect because of the added I^- from NaI. According to Le Chatelier (and the Ksp expression) this will reduce the solubility of PbI₂ so [Pb²⁺] will be LESS than in part (a)

(c) PbCl₂ will have the highest concentration of Pb²⁺ because it has the largest Ksp meaning that the molar solubility of Pb²⁺ will be 2x10^-5 M compared to the others where the molar solubility of Pb²+ will be only 7x10^-9 M or 3x10^-13 M